# Whole roots and powers

1. Nov 28, 2011

### georg gill

$$(a^{\frac{1}{2n}}a^{\frac{1}{2n}})^n=(a^{\frac{1}{2n}})^n(a^{\frac{1}{2n}})^n=(a^{\frac{1}{2}}) (a^{\frac{1}{2}})=a=(a^{\frac{1}{n}})^n$$

all above is just done by using that the order of the factors that you multiply does not matter

we have proven that

$$(a^{\frac{1}{p}}a^{\frac{1}{p}})^n=(a^{\frac{2}{p}})^n$$

for any even number. Try saying that p is a odd number then n is a decimal number and taking the rooth with a number that is not whole does not make sense in a logixal approach of what one can really comprehend. For any even number of p what I did above makes sense. Could someone see an extention of this system. I want to make left side here:

$$\sqrt[m]{\underbrace{a^{\frac{1}{n}} \, \cdot \, a^{\frac{1}{n}} \, \cdot \, . . . \, \cdot a^{\frac{1}{n}} \, }_{\text{m times}} \, }=a^{\frac{m}{n}}$$

to become right side by only using roths and powers of integers.

Last edited: Nov 28, 2011
2. Nov 28, 2011

### HallsofIvy

Certainly $a^{1/2n}a^{1/2n}= a^{1/2n+ 1/2n}= a^{2/2n}= a^{1/n}$.

And $\left(a^{1/n}\right)^m= a^{(1/n)m}= a^{m/n}$. Those are well known properties of the exponentials.

Last edited by a moderator: Nov 28, 2011
3. Nov 28, 2011

### D H

Staff Emeritus
If you know what Dedekind cuts are, the PlanetMath page http://planetmath.org/encyclopedia/ProofOfPropertiesOfTheExponential.html [Broken] does a nice job of extending exponentiation from the integers to the rationals, and then to the reals.

Last edited by a moderator: May 5, 2017
4. Nov 29, 2011

### georg gill

As far as I can see or from what I get from the proof at least

$$(xy)^m)^{\frac{1}{n}}=(y)^m)^{\frac{1}{n}}(x)^m)^{\frac{1}{n}}$$

I want to prove that

$$(xy)^m)^{\frac{1}{n}}=y)^{\frac{m}{n}}x^{\frac{m}{n}}$$

What I have managed to show is that

$$(xy^{\frac{1}{np}})^{np}$$

$$(a^{\frac{1}{np}})^{np}=a$$

$$((a^{\frac{1}{n}})^{\frac{1}{p}})^{np}=(((a^{\frac{1}{n}})^{\frac{1}{p}})^n)^p$$

since n and p are integers to put p outside as own power goues without proving (could of course prove this
and we get

$$((a^{\frac{1}{n}})^{\frac{1}{p}})^{np}=(((a^{\frac{1}{n}})^{\frac{1}{p}})^n)^p=a$$

so

$$xy^{\frac{1}{np}}=(a^{\frac{1}{n}})^{\frac{1}{p}}$$

Last edited by a moderator: May 5, 2017